That's an interesting point and I suppose reversibility, interpreted as bijectivity, is not enough.
But unitarity implies a lot more. More specifically, unitary operators preserve distance. So we can prove a short impossibility result on limits existing, i.e., $U^n\vert x\rangle$ can never converge unless $\vert x\rangle$ is already an eigenvector of eigenvalue $1$.
Suppose for a contradiction that $U^nx$ converges to some state $y$ (I won't use kets just to make notation easier). Let $\epsilon > 0$. Pick whatever $N_1$ is large enough such that $\Vert U^N x - y\Vert < \epsilon$ for all $N\geq N_1$. But notice that $U^{N_1}$ is also unitary, so we can argue that
$$ \Vert U^{N_1}x - U^{2N_1}x\Vert = \Vert U^{N_1}(x - U^{N_1}x)\Vert = \Vert x - U^{N_1}x\Vert$$
and then use the triangle inequality
$$ \Vert U^{N_1}x - U^{2N_1}x\Vert \leq \Vert U^{N_1}x - y\Vert + \Vert y - U^{2N_1}x\Vert < 2\epsilon$$
but then we can further argue that
$$ \Vert x - y\Vert \leq \Vert x - U^{N_1}x\Vert + \Vert U^{N_1}x - y\Vert < 3\epsilon$$
by combining the equation and inequality above. Since $\epsilon$ was arbitrary, this implies $x=y$.